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Put $D=\mathbb{Q}/\mathbb{Z}$. The character module $M^{\ast }={\mathrm{Hom}}_{\mathbb{Z}}(M,D)$ has the natural actionAdditivity is immediate, and $a(b\varphi )(m)=\varphi (bam)=\varphi (abm)$, using commutativity of $A$. The identity of $A$ acts as the identity. We establish the extension and duality tools used in the four implications, so no flatness or injectivity criterion is assumed without proof.

An injective module $E$ is one for which every homomorphism from a submodule $U\subseteq V$ extends to $V$. The Baer criterion says it suffices to test the inclusions of ideals $I\subseteq A$. Necessity follows directly from the definition. For sufficiency, suppose every $I\to E$ extends to $A$. Given $h:U\to E$, use Zorn's lemma to choose a maximal extension $(W,h_W)$ inside $V$; extensions along a chain have a well-defined union. If $v\in V\setminus W$, letThis is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Thenis well-defined: if two expressions represent the same element, their coefficient difference lies in $I$, where the two prescribed values agree. It extends $h_W$ to the strictly larger module $W+Av$, a contradiction. Thus $W=V$, proving the Baer criterion for arbitrary modules and ideals.

The group $D$ is divisible. A homomorphism $n\mathbb{Z}\to D$, for $n>0$, extends to $\mathbb{Z}$ by choosing $d\in D$ with $nd$ equal to its value at $n$; the zero ideal causes no restriction. The proved Baer criterion over $\mathbb{Z}$ therefore makes $D$ injective. It is also a cogenerator: a nonzero element $x$ of an abelian group generates either a finite cyclic group of order $n>1$, which maps $x$ to $1/n+\mathbb{Z}$, or an infinite cyclic group, which maps $x$ to $1/2+\mathbb{Z}$. Injectivity extends this character to the whole group. Thus every nonzero element is detected by an additive character.

It follows that a group homomorphism $u:T\to U$ is injective exactly whenis surjective. For an injection, extension into the injective group $D$ proves surjectivity. Conversely, a character nonzero on an element of $\ker u$ cannot be the pullback of a character on $U$. This is the detection property of the injective cogenerator of abelian groups.

Finally, for any $A$-module $N$ there is a natural character form of the Tensor-hom adjunction:An $A$-linear map $h$ gives the pairing $(n,m)\mapsto h(n)(m)$. It is balanced because $h(an)(m)=(ah(n))(m)=h(n)(am)$. Conversely a balanced additive pairing determines $h$ by this formula. The universal property of the tensor product of modules makes these constructions mutually inverse, and compatible with restriction maps.

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

We prove the forward direction of the finite-generator criterion for a natural uniform algebra. Suppose that $A$ is a natural uniform algebra, and let $f_1,\dots ,f_n$ have no common zero on $K$. The idealis all of $A$. Otherwise it is contained in a maximal ideal $M$, whose associated character of an algebra vanishes on all the $f_j$. Since every algebra character is an evaluation character under naturality, this would give a common zero, a contradiction. Therefore $1\in J$, meaning that

Here are the Banach-algebra details behind the algebra character associated with $M$. A proper ideal cannot have dense closure: if it contained an element within distance less than one of $1$, the Neumann series would make that element invertible, forcing the ideal to contain $1$. Thus the closure of a proper ideal is proper. A maximal ideal, obtained from Zorn's lemma, is therefore closed. Its quotient is a complex unital Banach algebra in which every nonzero element is invertible. The Gelfand-Mazur theorem identifies this quotient with $\mathbb{C}$, so the quotient map is the required algebra character. No closedness of the originally generated ideal $J$ is needed.

Take $\mathbb{N}=\{1,2,\dots \}$. We first prove the Ellis–Numakura lemma in the one-sided continuity convention needed here. Let $S$ be a nonempty compact Hausdorff space with an associative multiplication for which each map $x\mapsto xa$ is continuous. Among its nonempty closed subsemigroups, there is an inclusion-minimal one, $M$: a decreasing chain has nonempty intersection by compactness, and that intersection is a closed subsemigroup, so Zorn's lemma applies.

Fix $a\in M$. The set $Ma$ is nonempty and compact, hence closed in the Hausdorff space, and is a subsemigroup becauseMinimality gives $Ma=M$. Thus the set $T=\{x\in M:xa=a\}$ is nonempty. It is closed by the stated continuity, and it is a subsemigroup, since $(xy)a=x(ya)=xa=a$ for $x,y\in T$. Minimality again gives $T=M$, so $a\in T$ and $\boxed{a^2=a}$, giving a semigroup idempotent. If one instead uses the opposite one-sided continuity convention, the same proof uses $aM$ and $\{x:ax=a\}$; no joint continuity is required.

Apply this lemma to the Stone-Čech compactification of the natural numbers with its given addition. In the addition on the Stone-Čech compactification of the natural numbers convention,The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply there exists an idempotent ultrafilter $p$ with $p+p=p$.

On the positive integers such a $p$ is nonprincipal: a principal ultrafilter at $n$ adds to itself to give the one at $2n$, which is different. Hence every cofinite set belongs to $p$. If zero is included in one's convention for $\mathbb{N}$, use the same compact semigroup argument on the closed space of nonprincipal ultrafilters: it is nonempty by compactness of the infinite discrete set's compactification, and the displayed addition sends two free ultrafilters to a free ultrafilter. This avoids the trivial principal idempotent at zero.

To deduce Hindman's theorem, let $A$ be the unique colour class belonging to $p$. DefineIdempotence gives $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma also gives $A^{\ast }-n\in p$ whenever $n\in A^{\ast }$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ givesIntersecting this set with $A-n$ produces exactly $A^{\ast }-n$.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_r$ have been chosen with every nonempty finite sum in $A^{\ast }$. The finite intersectionbelongs to $p$, and is therefore nonempty. Choose $x_{r+1}$ from it. The new sums are $x_{r+1}$ and $s+x_{r+1}$, and they all lie in $A^{\ast }$. Induction yieldsThis proves every finite colouring of the positive integers admits a monochromatic finite-sums set generated by a strictly increasing infinite sequence, the required Hindman theorem.

We use Zorn's lemma, and no basis or other existence theorem from linear algebra. A subset $B\subseteq V$ is linearly independent if every finite relation ${\sum }_{b\in F}{\lambda }_{b}b=0$ forces every coefficient to be zero.

Order the linearly independent subsets by inclusion. This partially ordered set is nonempty, because the empty subset is independent. An empty chain has the empty subset as an upper bound. For any nonempty chain, its union is independent: every finite subset of that union lies in one member of the chain. Indeed choose one chain member containing each of its finitely many vectors, then take the largest among that finite collection of comparable members. Every finite relation in the union is therefore a relation in an independent chain member. The union is an upper bound.

Zorn's lemma gives a maximal independent subset $B$. If $v$ were outside its span, then $B\cup \{v\}$ would still be independent. In a finite relationa nonzero $\lambda$ could be inverted in the field $k$, placing $v$ in the span of $B$. If $\lambda =0$, independence of $B$ forces the remaining coefficients to be zero. This contradicts maximality, so $B$ spans $V$.

Define the free module on $B$ as the direct sum $k^{(B)}={⨁}_{b\in B}k$: its elements are coefficient families with finite support. The mapis a module homomorphism. Spanning proves surjectivity; independence proves injectivity. Hence the vector-space freeness from maximal independence givesFinite support is essential; this is a direct sum, not an unrestricted product. The zero vector space uses $B=\varnothing$ and is free of rank zero. The proof for arbitrary vector spaces explicitly uses the axiom of choice through Zorn's lemma.

Put $\kappa =|X|$, an infinite cardinal number. The original PDF has the double exponent $2^{2^{|X|}}$; the single-exponent expression in the TeX aid is a transcription error.

We explicitly construct an independent family of subsets with $2^{\kappa }$ members on a set of size $\kappa$. LetFor each finite $s$ there are finitely many possibilities for $H$. The collection of finite subsets of an infinite cardinal has size $\kappa$, so $|D|\le \kappa$; the points $(\{\xi \},\varnothing )$ give the reverse bound. Thus $|D|=\kappa$.

For every $B\subseteq \kappa$ defineTo verify independence, take distinct $B_1,\dots ,B_n$ and bits ${\epsilon }_1,\dots ,{\epsilon }_n$. For each pair $i<j$, choose a point of $B_i△B_j$ and let $s$ contain all these finitely many points. The traces $B_i\cap s$ are pairwise distinct. SetThen $(s,H)\in A_{B_i}$ exactly for those $i$ with ${\epsilon }_i=1$. Every finite Boolean pattern is therefore realized. The cases $n=0,1$ work by the same definition, allowing $s=\varnothing$. In particular the sets $A_B$ are all distinct, so the finite-trace independent family construction gives exactly $2^{\kappa }$ independent members.

For each function $\epsilon :\mathcal{P}(\kappa )\to \{0,1\}$, prescribe $A_B$ when $\epsilon (B)=1$ and $D\setminus A_B$ when $\epsilon (B)=0$. Independence gives the finite intersection property. Their finite intersections generate a proper filter on a set: its members are the subsets of $D$ containing one of those intersections.

To extend this to an ultrafilter, order its proper filter extensions by inclusion. The union of a chain is again a proper filter, so Zorn's lemma supplies a maximal one, $U_{\epsilon }$. If neither a subset $Y$ nor its complement belonged to this filter, adjoining $Y$ would still generate a proper filter: otherwise some existing member would be disjoint from $Y$ and force its complement into the filter already. Maximality therefore gives the ultrafilter dichotomy. Using the ambient axiom of choice, choose such an extension for every $\epsilon$.

Distinct functions differ at some $B$. Their ultrafilters respectively contain $A_B$ and its complement, and cannot be equal. There are $2^{2^{\kappa }}$ such functions. Transporting the ultrafilters along a bijection $D\to X$ proves the lower bound. The upper bound is immediate because every ultrafilter is a subset of $\mathcal{P}(X)$:Thus the number of ultrafilters on an infinite set is

A prime ideal $\mathfrak{p}$ is minimal over $I$ if $I\subseteq \mathfrak{p}$ and no strictly smaller prime contains $I$. An associated prime of a module is an annihilator of an individual nonzero element which happens to be prime. For the quotient module this readswhere ${\mathrm{Ann}}_R(\bar{a})=(I:a)=\{r\in R:ra\in I\}$. This is the annihilator of an individual element, rather than necessarily the annihilator of the whole quotient.

Minimal primes exist because $I$ is proper. First choose a maximal ideal containing $I$. Within the primes contained in it and containing $I$, an intersection of any decreasing chain is again prime. Indeed, if $ab$ is in the intersection and $a$ is absent from one member, then $b$ belongs to that member and to every smaller member; it also belongs to every larger member. The intersection is still proper and contains $I$. Zorn's lemma, applied with reverse inclusion, produces a minimal prime. This proves existence of minimal primes over a proper ideal, even without the Noetherian hypothesis.

Now fix a minimal prime $\mathfrak{p}$ and put $A=R/I$. The localization at a prime ideal $A_{\mathfrak{p}}$ is nonzero and is a Noetherian local ring. By prime ideal correspondence for localization, its only prime is $\mathfrak{p}{R}_{\mathfrak{p}}/I{R}_{\mathfrak{p}}$. Write this maximal ideal as $\mathfrak{n}$. It is the nilradical; since it is finitely generated and each generator is nilpotent, some power of $\mathfrak{n}$ is zero. Explicitly, if its generators have nilpotence exponents $e_1,\dots ,e_r$, every product of degree $1+{\sum }_i(e_i-1)$ vanishes.

Choose the smallest $s\ge 1$ such that ${\mathfrak{n}}^s=0$, and choose a nonzero element in ${\mathfrak{n}}^{s-1}$; when $\mathfrak{n}=0$, choose $1$. Its annihilator over $R_{\mathfrak{p}}$ is exactly $\mathfrak{p}{R}_{\mathfrak{p}}$. Represent it as $a/u$ with $u\notin \mathfrak{p}$. Multiplication by the unit $u$ shows that $a/1$ has the same annihilator and remains nonzero.

Let $p_1,\dots ,p_r$ generate $\mathfrak{p}$ in $R$. For each $i$, the equality $p_{i}a/1=0$ supplies $u_i\notin \mathfrak{p}$ such that $u_i{p}_i\bar{a}=0$ in $A$. Put $u={\prod }_i{u}_i$ and $\bar{b}=u\bar{a}$. Its localization is nonzero, so $\bar{b}\ne 0$, and every element of $\mathfrak{p}$ kills it. Conversely, an element outside $\mathfrak{p}$ becomes a unit and cannot kill the nonzero element $\bar{b}/1$. ThereforeThe key step in minimal primes are associated primes is clearing denominators for a finite generating set of $\mathfrak{p}$; that is where Noetherianity is used.

For an embedded associated prime, take $R=k[x,y]$ and $I=(x^2,xy)$. Its radical is $(x)$, so $(x)$ is its unique minimal prime. The nonzero class of $x$ satisfiesIndeed, $rx\in x(x,y)$ is equivalent to $r\in (x,y)$ by cancellation in the polynomial domain. Thus $(x,y)$ is associated but is not minimal, since $(x)⊊(x,y)$. For comparison, ${\mathrm{Ann}}_R(\bar{y})=(x)$, exhibiting the minimal associated prime as well.

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.

The bounding number $\mathfrak{b}$ is the least size of an unbounded family in $({\omega }^{\omega },{\le }^{\ast })$. The almost disjointness number $\mathfrak{a}$ is the least size of an infinite maximal almost disjoint family on omega of infinite subsets of $\omega$. Requiring the family to be infinite excludes trivial finite maximal partitions.

A countable family $\{f_i:i<\omega \}$ is bounded by $g(n)=1+{\max }_{i\le n}{f}_i(n)$. Thus ${\aleph }_1\le \mathfrak{b}$.

Suppose an infinite almost disjoint family on omega $\mathcal{A}$ has size less than $\mathfrak{b}$. Choose distinct members $A_n$ and put $C_n=A_n\setminus {\bigcup }_{i<n}{A}_i$. These are infinite and pairwise disjoint. For each $B\in \mathcal{A}$, define $h_B(n)=\max ((B\cap C_n)\cup \{0\})$ whenever the intersection is finite. If $B=A_n$ at the exceptional index $n$, set $h_B(n)=0$. These are the only possible infinite intersections. A single $g$ eventually dominates every $h_B$, because there are fewer than $\mathfrak{b}$ of them. Pick $x_n\in C_n$ with $x_n>g(n)$ and put $X=\{x_n:n<\omega \}$.

For $B$ not among the selected $A_n$, only finitely many $x_n$ can lie in $B$. The same is true for $B=A_j$, with its single exceptional index $j$ ignored. Thus $X$ is infinite and almost disjoint from every member of $\mathcal{A}$, so the family was not maximal. This bounding-to-almost-disjointness inequality proves $\mathfrak{b}\le \mathfrak{a}$.

Finally start with an infinite pairwise disjoint family and use Zorn's lemma to extend it to a maximal almost disjoint family on omega. It is a family of subsets of $\omega$, so has cardinality at most $2^{{\aleph }_0}$. Therefore

Use the Hausdorff convention for a locally convex space: a real vector space $X$ is equipped with a separating family $\mathcal{P}$ of seminorms. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\ne 0$ some $p\in \mathcal{P}$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting ofEquivalently, it is the coarsest vector-space topology making all these seminorms continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous linear functionals could not distinguish points in the common kernel of the seminorms.

The continuous dual space $X^{\ast }$ consists of all continuous linear functionals $X\to \mathbb{R}$. We first prove the needed Hahn-Banach theorem. Let $p$ be a real-valued sublinear function on $X$, meaning $p(a+b)\le p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\ge 0$, and let $g$ be linear on a vector subspace $M$, with $g(m)\le p(m)$. To extend across $v\notin M$, writeThe necessary bounds on $c$ areThey are compatible becauseTaking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in [L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\tilde{g}\le p$ on $M+\mathbb{R}v$. This is the one-dimensional dominated extension of a real linear functional.

Order all dominated extensions of $g$ by extension of their domains. A chain has an upper bound obtained by taking the union of the domains and linear functionals. Zorn's lemma gives a maximal extension, and the one-dimensional construction shows that its domain must be all of $X$. We have therefore proved the real dominated-extension theorem. In particular, when $p$ is a seminorm, domination at both $x$ and $-x$ gives $|f(x)|\le p(x)$.

For $x_0\ne 0$, choose a continuous seminorm $p$ with $p(x_0)>0$, and define $g(t{x}_0)=tp(x_0)$ on its one-dimensional span. Then $|g(t{x}_0)|\le p(t{x}_0)$. Extend by the theorem just proved to $f$ with $|f(x)|\le p(x)$. This bound makes $f$ continuous, and $f(x_0)=p(x_0)>0$. Applying this to the difference of two distinct points proves that $X^{\ast }$ separates the points of $X$, the continuous-dual separation theorem for Hausdorff locally convex spaces.

For separation from a closed linear subspace, choose a basic balanced neighborhood $V$ of zero such that $(x_0+V)\cap Y=\varnothing$. WriteThen $q(x_0-y)\ge 1$ for every $y\in Y$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$, which is well-defined since $x_0\notin Y$. For $t\ne 0$,for $t=0$ the inequality is immediate. The proved extension theorem gives a continuous $f$ dominated by $q$, withThis proves separation of a point from a closed linear subspace without using any unproved Hahn-Banach extension or separation result.

A proper filter on a set $\mathcal{F}$ on $\mathbb{N}$ contains $\mathbb{N}$, excludes the empty set, is closed under finite intersections and is upward closed under inclusion. An ultrafilter is a maximal proper filter. Equivalently, for every $A\subseteq \mathbb{N}$ it contains exactly one of $A,A^c$. To see maximality implies this dichotomy, if $A$ is absent then adjoining it must make the generated filter improper; hence some filter member is disjoint from $A$, forcing $A^c$ into the filter. Conversely, a filter with this dichotomy cannot be properly enlarged without acquiring disjoint members.

Extend the cofinite filter to a maximal proper filter using Zorn's lemma. Every chain of proper extensions has its union as a proper filter upper bound: any finite collection of its members lies in one member of the chain, and the empty set never enters. The resulting ultrafilter contains no finite set, because it already contains that set's cofinite complement. It is therefore a free ultrafilter, or nonprincipal ultrafilter. This proves the required existence with the usual choice principle explicit.

Define the Stone-Čech compactification of the natural numbers as the set of all ultrafilters, with basic open setsThey form a basis because $\hat{\mathbb{N}}$ is the whole space and $\hat{A}\cap \hat{B}=\widehat{A\cap B}$. Also $(\hat{A}{)}^c=\widehat{A^c}$, so each basic set is clopen. Distinct ultrafilters disagree on some $A$, putting them in the disjoint open sets $\hat{A}$ and $\widehat{A^c}$. Thus this space is Hausdorff.

To prove compactness, suppose an open cover has no finite subcover and refine it by basic sets $\widehat{A_i}$. Their complements $B_i=\mathbb{N}\setminus A_i$ have the finite intersection property: otherwise finitely many $\widehat{A_i}$ would cover every ultrafilter. Here a nonempty finite intersection of the $B_i$ has a principal ultrafilter containing it, whereas an empty intersection cannot belong to any proper filter. The $B_i$ therefore generate a proper filter, which extends to an ultrafilter by the same Zorn argument. That ultrafilter lies outside every $\widehat{A_i}$, contradicting the cover. Hence $\beta \mathbb{N}$ is compact and Hausdorff. Natural number $n$ is identified with its principal ultrafilter.

Hindman's theorem asserts that every finite colouring of the positive integers has an increasing sequence $x_1<x_2<\cdots$ for which all nonempty finite sums of distinct terms have one colour. Let $\mathcal{U}+\mathcal{U}=\mathcal{U}$ be the given idempotent ultrafilter. Since $\mathbb{N}$ is positive, a principal ultrafilter cannot be idempotent: its sum with itself is principal at $2n$, not $n$. Thus $\mathcal{U}$ is nonprincipal and contains every cofinite tail. In a convention including zero, one instead uses an idempotent in the nonprincipal part, excluding the trivial principal idempotent at zero.

For $A\subseteq \mathbb{N}$ define $A-n=\{m:n+m\in A\}$. Addition on the Stone-Čech compactification of the natural numbers is characterized byOne cell $A$ of the finite colour partition belongs to $\mathcal{U}$. PutIdempotence makes $A^{\ast }\in \mathcal{U}$. Moreover, if $n\in A^{\ast }$, then $A^{\ast }-n\in \mathcal{U}$. Indeed, write $B=\{y:A-y\in \mathcal{U}\}$ so $A^{\ast }=A\cap B$. We have $A-n\in \mathcal{U}$, and idempotence applied to that set givesTheir intersection is $A^{\ast }-n$. This is the idempotent-ultrafilter star-set lemma.

Choose $x_1\in A^{\ast }$. If the finite-sums set $F_k$ of the first $k$ choices lies in $A^{\ast }$, selectEvery factor belongs to $\mathcal{U}$, so this finite intersection is nonempty. Its choice preserves $F_{k+1}\subseteq A^{\ast }$. ThereforeThis proves the requested Idempotent-ultrafilter proof of Hindman's theorem, without assuming an idempotent-existence proof.

Finally, for the first logical assertion put $P=\{x:p(x)\}$ and $Q=\{x:q(x)\}$. A proper filter contains $P\cap Q$ if and only if it contains both $P$ and $Q$: one direction is finite-intersection closure and the other is upward closure. Thus (i) is always true. This is the conjunction law for filter quantifiers.

A Noetherian ring is a ring whose ideals satisfy the ascending chain condition; equivalently every ideal is finitely generated. For the surjective ring endomorphism, put $K_j=\ker f^j$. The ascending chain of ideals $K_1\subseteq K_2\subseteq \cdots$ stabilizes, say $K_m=K_{m+1}$. If $a\in \ker f$, surjectivity of $f^m$ gives $b$ with $f^m(b)=a$. Then $f^{m+1}(b)=0$, so $b\in K_{m+1}=K_m$ and $a=0$. The endomorphism is therefore injective, hence a ring automorphism.

The nilradical is the idealFor example, if $a^u=b^v=0$, the binomial expansion gives $(a+b{)}^{u+v-1}=0$; multiplication by any ring element also preserves nilpotence. Every prime ideal contains each nilpotent element, since $a^j\in P$ implies $a\in P$. Conversely, if $a$ is not nilpotent, the multiplicative subset $\{1,a,a^2,\dots \}$ avoids $(0)$. By Zorn's lemma, choose an ideal maximal among those disjoint from this subset. It is a prime ideal: if $xy\in J$ but neither factor is in $J$, then $J+(x)$ and $J+(y)$ both meet the subset, and multiplying their two witnesses puts a power of $a$ in $J$, a contradiction. This prime ideal excludes $a$. HenceFor the zero ring the empty intersection is the whole ring, as required.

Here is a finite prime-intersection representation of a radical argument that does not assume the result. In a Noetherian ring, suppose some ideal $J$ has ideal radical not expressible as a finite intersection of prime ideals, and choose such a $J$ maximal by the ascending chain condition. It is proper and cannot be prime. Choose $a,b\notin J$ with $ab\in J$. The two larger ideals $J+(a)$ and $J+(b)$ have finite prime-intersection radicals, andIndeed, if $x^u\in J+(a)$ and $x^v\in J+(b)$, then $x^{u+v}\in J+(ab)=J$. This contradicts the choice of $J$. Applying the result to $(0)$ proves that $N(R)$ is the intersection of finitely many prime ideals. Empty intersections handle the unit ideal.

For reducedness of a formal power series ring, the inclusion of constants gives one implication: a nonzero nilpotent element of $R$ remains nonzero and nilpotent in $R[[X]]$. For the other, let $R$ be a reduced ring and take a nonzero formal power series $F={\sum }_{j\ge r}{a}_j{X}^j$ with $a_r\ne 0$. The coefficient of $X^{mr}$ in $F^m$ is $a_r^m\ne 0$, because $R$ is reduced. Thus $F$ is not nilpotent. ThereforeNo Noetherian ring hypothesis is needed for this argument.

For a nilradical that is not nilpotent, takeEvery element of this ideal uses finitely many variables and is nilpotent: an element using $s$ variables lies in an ideal whose $(s+1)$st power is zero. The quotient by this ideal is the field $k$, so it is exactly the nilradical. Nevertheless ${\bar{t}}_1\cdots {\bar{t}}_m\ne 0$ for every $m$, because the square-free monomials form a vector space basis. Thus $N(R{)}^m\ne 0$ for all $m$.

The completeness theorem for propositional logic states that $S\models \varphi$ if and only if $S⊢\varphi$. Soundness follows by checking that axioms are tautologies and inference rules preserve truth. For completeness it suffices to prove that every syntactically consistent set has a model.

The deduction theorem for propositional logic used here is: $S\cup \{\psi \}⊢\chi$ if and only if $S⊢\psi \Rightarrow \chi$, in classical propositional logic. A proof is finite, so the union of a chain of consistent sets remains consistent: a contradiction would already occur in one member. By Zorn's lemma, extend a consistent $S$ to a maximal consistent set $T$. If $\psi \notin T$, adding $\psi$ yields a contradiction; the deduction theorem gives $T⊢\neg \psi$, and maximal consistency puts $\neg \psi$ in $T$. Conversely $\psi$ and $\neg \psi$ cannot both belong to $T$. Maximal consistency also makes $T$ deductively closed.

Assign each primitive proposition the truth value true precisely when it belongs to $T$. The maximal consistent set truth lemma follows by induction on formulas. For negation use the preceding dichotomy. For implication, if $\neg \psi \in T$ or $\chi \in T$, the classical implication axioms put $\psi \Rightarrow \chi$ in $T$; if $\psi \in T$ and $\neg \chi \in T$, modus ponens prevents that implication from belonging to $T$. This is exactly its truth table. Other connectives are defined from these or handled by the same induction. The resulting valuation satisfies all of $T$, hence all of $S$.

If $S⊬\varphi$, the deduction theorem and classical contradiction rules show $S\cup \{\neg \varphi \}$ is consistent. Its model falsifies $\varphi$, so $S/\models \varphi$. This proves completeness.

The propositional compactness theorem says $S$ is satisfiable if every finite subset is satisfiable. Otherwise completeness would make $S$ inconsistent, and a finite proof of contradiction would involve only finitely many assumptions, contradicting their satisfiability. Equivalently, every semantic consequence of $S$ is a consequence of a finite subset. The decidability theorem for propositional logic says there is an algorithm deciding whether a formula is provable with no assumptions: enumerate all truth assignments to its finitely many primitive propositions, check whether it is a tautology, and use soundness and completeness to identify tautologies with theorems. This does not claim an algorithm for arbitrary infinitely presented theories.

Finally, the valuation assigning false to every $p_n$ satisfies all the implications in $S$ and falsifies $p_1$. By soundness, $S⊬p_1$.

Assume the axiom of choice and suppose a poset satisfying the hypothesis has no maximal element. Fix a choice function on its nonempty subsets. By transfinite recursion, choose at each ordinal stage an upper bound of the chain in a partial order already constructed, then choose a strictly larger member. Both choices are available, including the initial empty stage. This produces distinct members throughout the Hartogs ordinal of $P$, contradicting the Hartogs theorem. Thus a maximal element exists. Conversely, apply Zorn's lemma to partial choice functions on a family of nonempty sets, ordered by extension. A chain in a partial order has the compatible union as an upper bound, and a maximal element must be defined on the whole family, since one missing value could be adjoined.

Zorn's lemma was introduced in 1935 to shorten algebra proofs using choice or the well-ordering theorem.^[ref]

If $p$ is a sublinear functional, $q$ a superlinear functional and $q\le p$ on a real vector space, there exists a linear functional $U$ between them. More generally, a linear functional $S$ on a subspace extends with this domination whenever $S(y)\le p(x+y)-q(x)$ for all ambient $x$ and subspace $y$. In a one-dimensional extension the admissible values for a sandwich extension form a nonempty real interval. Positive homogeneity checks the extension on both signs of the new coordinate. Chain unions preserve linearity and the mixed domination, so Zorn's lemma gives a maximal extension, whose domain must be the whole space. Finally $x=0$ in the mixed inequality gives the upper bound; using the pair $(x,-x)$ gives the lower bound. Conversely, any linear $U$ between $q,p$ obeys the mixed inequality by $U(y)=U(x+y)-U(x)$.

Every partial order has a total order extension. Apply Zorn's lemma to extensions ordered by inclusion: the union of a chain is a partial order. If a maximal extension leaves two elements incomparable, adjoining one comparison and taking its transitive closure of a relation creates no cycle, contradicting maximality. Quotienting a preorder by its indifference relation and lifting a total order gives a complete weak-order extension preserving its strict part.

Every local subring $A$ of a field $K$ is dominated by a valuation ring with fraction field $K$. Order local overrings by inclusion and contraction of their maximal ideals; chain unions are local, so Zorn's lemma gives a maximal one. For each $z\in K^{\times }$, at least one of the extensions of its maximal ideal to $V[z]$ and $V[z^{-1}]$ is proper: otherwise two least-degree relations for $1$ with maximal-ideal coefficients reduce one another to a smaller-degree relation. Localizing the proper extension gives a dominating overring, so maximality forces either $z$ or $z^{-1}$ into $V$.